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Question Number 103219 by Dwaipayan Shikari last updated on 13/Jul/20

1 + 1 + 1 + 1 + 1 + 1 + 1 + \dots . = S_{n}

S_{n} = 1 + 1 + 1 + 1 + 1 + 1 + 1 + \dots .

2 S_{n} = 2 + 2 + 2 + \dots \dots .

\dots \dots \dots . s u b t r a c t i n g

- S_{n} = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + \dots .

- S_{n} = \frac{1}{2}

S_{n} = - \frac{1}{2}

I h a v e f o u n d t h i s w h i l e e x p e r i m e n t . I k n o w t h e s u m d i v e r g e s

b u t i s i t p r e t t y c o o l ?

K i n d l y r e c t i f y m e i f t h e r e i s a n y f a u l t o n t h i s n o n r i g o r o u s

p r o c e s s

I h a v e f o u n d s o m e R a m a n u j a n p r o o f

S_{n} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + \dots

4 S_{n} = 4 + 8 + 12 + \dots

- 3 S_{n} = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots \dots

- 3 S_{n} = \frac{1}{4}

S_{n} = - \frac{1}{12}

R a m a n u j a n h a d d o n e t h i s o n h i s n o t e b o o k

Commented by JDamian last updated on 13/Jul/20

I a l w a y s h a v e b e e n t o l d \infty - \infty i s

u n d e f i n e d .

I n t h i s c a s e S_{n} = \infty \Rightarrow 2 S_{n} = \infty \Rightarrow

S_{n} - 2 S_{n} = \infty - \infty

Commented by Dwaipayan Shikari last updated on 13/Jul/20

I k n o w w h a t d o y o u w a n t t o s a y s i r . B u t I f w e {d o n}^{'} t t a k e S_{n} a s

i n f i n i t e t h e n w h a t w i l l h a p p e n ?

Commented by PRITHWISH SEN 2 last updated on 13/Jul/20

First of all \infty is not a number so \infty - \infty = 0

or \infty + \infty = 2 \infty is useless . \infty is a conception it

is a symbol which represents the number which

is greater than any large number what you can imagine

∵ it is not any arithmetic number that is why

no arithematic operations (such as +, -, \div, \times)

can be done with \infty .

Now as far series is concerned the addition or

substraction of two series can be down only

when these two series converges to some finite

values .

Now there are large as well as smaller \infty

like for the set of integer and for the set of real

numbers the \infty of the former set (natural number set) is smaller then the

\infty of the later set (real number set) . I think this

much is enough for your doubt . Thank you .

Answered by prakash jain last updated on 13/Jul/20

ζ (0) = - \frac{1}{2}, ζ (- 1) = - \frac{1}{12}

ζ (s) i think you already know .

I think you should read the links that

I sent earlier on this topic to get

some insight on why it works .

There a quite a bit of maths involved

and I really dont want to retype

material available .

I admire your curiosity and if you

put some more effort you will feel

more comfortable when looking

at sum of divergent series such

as 1 + 2 + 3 + 4 + \dots = - \frac{1}{12}

Commented by prakash jain last updated on 13/Jul/20

I will add a simple example of

gemometric series

f (x) = \frac{1}{1 - x} = 1 + x + x^{2} + \dots ∣ x ∣< 1

using analytical continuity so

can sum

1 + 2 + 2^{2} + 2^{3} + . . = \frac{1}{1 - 2} = - 1

or you can write

S = 1 + 2 (1 + 2 + 2^{2} + . .) = 1 + 2 S

S = - 1

I am not adding any mathematical

proofs here they are given in the

links that i shared before .

Commented by Dwaipayan Shikari last updated on 13/Jul/20

I a m a l w a y s c u r i o u s a b o u t t h e s e t y p e o f s e r i e s .

Commented by Dwaipayan Shikari last updated on 13/Jul/20

Sir why two types of results are here

1 - 2 + 3 - 4 + 5 - 6 + \dots = \frac{1}{4}

but logically it approches to - \infty

or

1 - 2 + 3 - 4 + \dots . = - 1 - 1 - 1 - 1 - 1 - \dots = - (1 + 1 + 1 + 1 + 1 + 1 + . .)

= - (- \frac{1}{2}) = \frac{1}{2}

i {haven}^{'} t read the article fully

But it gives a overview that it is a arithmatic mean of some

results

I will read the full article

Thanking you sir for giving me this source .

Commented by Dwaipayan Shikari last updated on 13/Jul/20

This is link which sir gave to me I want to share with everyone https://en.m.wikipedia.org/wiki/Analytic_continuation